A hollow, thin-walled sphere of mass 14.0 kg and diameter 49.0 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by θ (t) = At2+Bt4, where A has numerical value 1.10 and B has numerical value 1.60.

Answer

Given,

θ (t) = At² + Bt⁴

where A = 1.10

B = 1.60

a) calculating the unit of A and B

Unit of A = unit of θ / unit of t²

= rad/s²

Unit of B = unit of θ / unit of t⁴

= rad/s⁴

(b)

calculating angular speed at time t = 3 s

mass m = 14 kg

diameter d = 49 cm

radius r = d / 2 = 24.5 cm = 0.245 m

Angular momentum of the sphere

L = I w

I = moment of inertia

I = (2 / 3) m r²

I = 0.667 x 14 x 0.245²

I = 0.5605 kg m²

angular speed

ω = dθ / dt

= 2At + 4Bt^3

= (2 x 1.10 x 3) + (4 x 1.6 x 3³)

= 179.4 rad / s

So, angular momentum

L = 0.5605 x 179.4

L = 100.55 kg m²/s

(c) now, calculating net torque

T = I α

α = angular acceleration = dω / dt

= 2 A + 12 B t²

= 2 x 1.1 + 12 x 1.6 x 3²

= 175 rad / s²

T = 175 x 0.5605

T = 98.08 N m